A hydrogen atom initially in the ground state absorbs a photon and is in the excited state with energy 12.5 ev
Answers
Answered by
0
For ground state n1=1 and n2=4
Energy of photon absorbed E=E2−E1
=13.6(112−142)
=13.6×1516ev
=13.6×1516×1.6×19−9
E=2.04×10−18J
λ=hcE=6.6×10−34×3×1082.04×10−18
λ=9.7×10−8m
ν=cλ
=3×1089.7×10−8
=3.1×10−15hz
Energy of photon absorbed E=E2−E1
=13.6(112−142)
=13.6×1516ev
=13.6×1516×1.6×19−9
E=2.04×10−18J
λ=hcE=6.6×10−34×3×1082.04×10−18
λ=9.7×10−8m
ν=cλ
=3×1089.7×10−8
=3.1×10−15hz
Similar questions